Another New Proof Method of Analytic Inequality

This paper gives another new proof method of analytic inequality involving n variables. As its applications, we give proof of some known-well inequalities and prove five new analytic inequalities. 1. monotonicity on special variables Throughout the paper R denotes the set of real numbers and R+ denotes the set of strictly positive real numbers. Let n ≥ 2, n ∈ N. The arithmetic mean A(x) and the power mean Mr(x) of order r with respect to the positive real numbers x1, x2, · · · , xn are defined respectively as A(x) = 1 n ∑n i=1 xi, Mr(x) = ( 1 n ∑n i=1 x r i )1/r for r 6= 0, and M0(x) = (∏ni=1 xi). In paper [4], the author puts up a new proof method of analytic inequality. In this section, we shall provide another new proof method of analytic inequality involving n variables. Lemma 1.1. Let interval I = [m,M ] ⊂ R, D def. = {(x1, x2) |m ≤ x2 ≤ x1 ≤M} ⊂ I2 and function f : I2 → R have continuous partial derivative. Then ∂f/∂x1 ≥ (≤) ∂f/∂x2 hold in D, if and only if f (a, b) ≥ (≤) f (a− l, b+ l) hold for all a, b ∈ I and l with b < b+ l ≤ a− l < a. Proof. Without the losing of generality, we only prove the case ∂f/∂x1 ≥ ∂f/∂x2. For all x1, x2 ∈ D and l ∈ R+ withm ≤ x2 < x2+l ≤ x1−l < x1 ≤M , we have f (x1 − l, x2 + l)− f (x1, x2) ≤ 0. Then it exists ξl ∈ (0, l) such that l ( − (x1 − ξl, x2 + ξl) ∂x1 + ∂f (x1 − ξl, x2 + ξl) ∂x2 ) ≤ 0, − (x1 − ξl, x2 + ξl) ∂x1 + ∂f (x1 − ξl, x2 + ξl) ∂x2 ≤ 0. Let l→ 0+, we get ∂f (x1, x2) ∂x1 ≥ ∂f (x1, x2) ∂x2 . According to continuity of partial derivative, we know ∂f (x1, x1) ∂x1 ≥ ∂f (x1, x1) ∂x2 .